Let g be defined on all of R. If A is a subset of R, define the set g^-1(A) by
g^-1(A)={x in R : g(x) in A}.
Show that g is continuous iff g^-1(O) is open whenever O contained in R is an open set.

well g^-1(O) means g(O) is in A.
Let g be continuous and O be an open subset of R.
Then |x-c|<delta and |g(x)-g(c)|<epsilon

Now I get stuck

Nov 28th 2010, 11:37 AM

nimon

"Then |x-c|<delta and |g(x)-g(c)|<epsilon."

This just won't do. Write back with the proper definition of continuity and we'll carry on from there.

Nov 28th 2010, 12:51 PM

kathrynmath

A function f:A--->R is continuous at a point c in A if, for all epsilon epsilon>0, there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.

Nov 28th 2010, 01:32 PM

kathrynmath

I tried to get started on both directions:

suppose g^-1(O) is open for each open subset O of R. Fix a real number x, and let epsilon be given. The neighborhood N centered at g(x) with radius epsilon is an open set, so the inverse image of N under g is an open set containing x. Thus there is a neighborhood V of x with radius delta such that V is a subset of g^-1(N)

If g is continuous at x, then there exists a delta such that whenever y is less than delta apart from x, g(y) is in a neighborhood of g(x) with radius epsilon.

Nov 28th 2010, 01:35 PM

Plato

Lets suppose that for each open set it is the case that is also open.
Given a point and then is an open set.
So is an open set that contains .
Hence, such that .
But .
Which by the definition is continuous.